applying Chinese Remainder Theorem Suppose  $N=p_1^{n_1}p_2^{n_2}...p_t^{n_t} $ where the $p_i$ are unique primes .
Now by the Chinese Remainder Theorem 
$$ SL_2(\mathbb{Z}/N\mathbb{Z})=SL_2(\mathbb{Z}/p_1^{n_1}\mathbb{Z}) \times SL_2(\mathbb{Z}/p_2^{n_2} \mathbb{Z}) \times...\times SL_2(\mathbb{Z}/p_t^{n_t}\mathbb{Z})$$ 
I only know that  $$ \mathbb{Z}/N\mathbb{Z}=\mathbb{Z}/p_1^{n_1}\mathbb{Z} \times \mathbb{Z}/p_2^{n_2}\mathbb{Z} \times... \times \mathbb{Z}/p_t^{n_t}\mathbb{Z}$$ by the 
Chinese Remainder Theorem  , but why is it also true for $SL_2(\mathbb{Z})=\lbrace
    \begin{pmatrix}
    a & b  \\
    c & d  \\
    \end{pmatrix}  :a,b,c,d \in \mathbb{Z} \ \ , ad-bc=1 \rbrace
$ ?
Thanks for help .
 A: Because, more generally, for any commutative rings $R$ and $S$, we have an isomorphism of matrix rings
$$\operatorname{M}_{n×n} (R × S) \cong \operatorname{M}_{n×n} R × \operatorname{M}_{n×n} S,$$
that restricts to isomorphisms of matrix groups
$$\operatorname{GL}_n (R × S) \cong \operatorname{GL}_n R × \operatorname{GL}_n S \quad\text{and}\quad \operatorname{SL}_n (R × S) \cong \operatorname{SL}_n R × \operatorname{SL}_n S.$$
The first isomorphism, the one of the matrix rings, should be fairly obvious: Just split up a matrix of component pairs into a pair of component matrices. No information is lost, addition and multiplication are preserved. Why does it restrict so nicely to these matrix groups?
It’s because $(R×S)^× = R^× × S^×$ and $(1,1)$ is the unit element in $R × S$ and the determinant of a matrix $A ∈ \operatorname{M}_{n×n} (R×S)$ is given by the determinants of its component matrices $A_R$ and $A_S$, that is
$$\det A = (\det A_R, \det A_S).$$
Hence, the isomorphism restricts to the general linear and then to the special linear group. You can remember this as “matrix functors respect products”.
